Advanced Math Help with Any Problems!

The velocity distribution for laminar flow between parallel plates is given by:
$\frac{u}{u_{max}}=1-<!--\; -\; -->{\left(\frac{2y}{h}\right)}^2$
where h is the distance separating the plates and the origin is placed midway between the plates. Consider a flow of water at ${15}^{\circ <!--\; \circ \; -->}C$ , with $u_{max}=0.30\frac{m}{sec}$ and $h=0.50mm$ . Calculate the shear stress on the upper plate and give its direction.

How to arrive at this integral?
However, I am not sure how to arrive at this:
$$\begin{array}{rl}\phantom{=}& {\int <!--\; \int \; -->}_0^1{x}^{n+2}(1-<!--\; -\; -->x{)}^{n}dx+2{\int <!--\; \int \; -->}_0^1{x}^{n+1}(1-<!--\; -\; -->x{)}^{n+1}dx+{\int <!--\; \int \; -->}_0^1{x}^n(1-<!--\; -\; -->x{)}^{n+2}dx\\ =& 2I(n+1)+2{\int <!--\; \int \; -->}_0^1{x}^{n+2}(1-<!--\; -\; -->x{)}^{n}dx\end{array}$$
and why the highlighted part is equal to each other? What type of integral is that?

Find a recursion formula for combinatorial problem
Let $C_n$ be the number of sequences with a length of n, which their elements belong to {0,1,2}, and they don't contain the following sequences: 11,21. Find a recursion formula with starting conditions for $C_n$.

What is the most general distribution for which $E[1/x]=1/E[x]$?
What is the most general distribution for which the expected value of the multiplicative inverse equals the multiplicative inverse of the expected value?
Motivation: I'm into modelling dynamics on graphs and I found a problem which is easily solvable in cases where the degree distribution of the vertices is a distribution where $E[1/k]=1/E[k]$. ($k_i$ is the degree of the ith vertex) From this solution I may gain an insight into how to unify multiple models.
So particularly I'm looking for a distribution which consists of non-negative, finite integers. But I'm also interested in continuous solutions. Distributions where $E[1/k^n]=1/E[k^n]$ may also help unifying the models.
What I do know so far, that $k_i=1$ is a particular solution. In the continuous case every function where $f(x)=f(1/x)$ and $E[x]=1$ is a solution. I know what momentum generating functions are and they seem like a good direction to try in, but I failed so far.
What is the most general form of this distribution? Does it have a name? It sounds like something trivial, like a "famous" distribution, but I can't find it.

Let $f:A\to <!--\; \to \; -->B$ and $g:C\to <!--\; \to \; -->D$. Define$$f\times <!--\; \times \; -->g=\{((a,c),(b,d)):(a,b)\in <!--\; \in \; -->f\text{ and }(c,d)\in <!--\; \in \; -->g\}.$$Prove that $f\times <!--\; \times \; -->g$ is a function from $A\times <!--\; \times \; -->C$ to $B\times <!--\; \times \; -->D$.
My proof (so far):
Let $(a,c)\in <!--\; \in \; -->A\times <!--\; \times \; -->C$, then $a\in <!--\; \in \; -->A$ and $c\in <!--\; \in \; -->C$.
Let $(b,d)\in <!--\; \in \; -->B\times <!--\; \times \; -->D$, $b\in <!--\; \in \; -->B$ and $d\in <!--\; \in \; -->D$.
If $p\in <!--\; \in \; -->A\times <!--\; \times \; -->C$ then $p=(a,c)$.
Define $(f\times <!--\; \times \; -->g)(p)$ as $(f\times <!--\; \times \; -->g)(p)=(f\times <!--\; \times \; -->g)(a,c)=(f(a),g(c))=(b,d)$, however $b\in <!--\; \in \; -->B$ and $d\in <!--\; \in \; -->D$ and $(b,d)\in <!--\; \in \; -->B\times <!--\; \times \; -->D$.
This shows $f\times <!--\; \times \; -->g$ is a function from $A\times <!--\; \times \; -->C$ to $B\times <!--\; \times \; -->D$ as $(f\times <!--\; \times \; -->g):A\times <!--\; \times \; -->C\to <!--\; \to \; -->B\times <!--\; \times \; -->D$ then $(a,b)\to <!--\; \to \; -->(f(a),g(c))$.
I know that it is not well written out, but I was wondering if my thought process so far made sense or if I accidentally missed something. Additionally, I am unsure of what the next step may be. As a sidenote, I am worried that this does not work for the given definition of $f\times <!--\; \times \; -->g$.

Extracting even / odd part of summation trick
Given a function f(x), we know its even part is given as $\frac{f(x)+f(-<!--\; -\; -->x)}{2}$ and its odd part is given by $\frac{f(x)-<!--\; -\; -->f(-<!--\; -\; -->x)}{2}$.
Consider a discrete sequence given by $a_j$ for $j\ge <!--\; \ge \; -->1$, then the sum of the terms in the sequence till n terms is given as
$$S=\underset{j}{\overset{n}{\sum <!--\; \sum \; -->}}{a}_j$$
Suppose I wanted to get the sum of the even terms in the above expression; then
$$S_{odd}=\underset{j}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{a_{-<!--\; -\; -->j}+a_j}{2}$$
and, for odd,
$$S_{odd}=\underset{j}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{a_j-<!--\; -\; -->a_{-<!--\; -\; -->j}}{2}$$
But wait, our sequence was defined for $j\ge <!--\; \ge \; -->1$. Well, here's the thing: $a_j$ is some function of j, extending the domain to negative integers and evaluating the function gives the right answer... but I can't understand why the continuous function trick extended to here.
Examples:
Sum of first n numbers given as $\frac{n(n+1)}{2}$, sum of first n odds will be given as: $\frac{n(n+1)}{2}-<!--\; -\; -->\frac{n(1-<!--\; -\; -->n)}{2}$
My attempt at finding an exact connection: To the discrete sequence $\frac{n(n+1)}{2}$, we can associate a function $f(x)=\frac{x(x+1)}{2}$ and we can think of the summation as summing this function at several different input points i.e:
$$S=\underset{j}{\overset{n}{\sum <!--\; \sum \; -->}}{a}_j\to <!--\; \to \; -->\underset{k}{\overset{n}{\sum <!--\; \sum \; -->}}f(x+k)$$
Then we apply the even odd decomposition and return back to the sequence world.
My question: Does there exist an association for every sequence with a function? If not, what is the criterion for an association to exist?